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Potential energy $U$ can be defined up to an arbitrary additive constant $c$ because $$F=-\dfrac{d(U+c)}{dx}=-\dfrac{dU}{dx}=ma$$ And therefore the equation of motion remains unchanged. I think the same holds for kinetic energy $T$ using a similar reasoning and I want to make sure that I'm getting it right.

In a system where conservation of mechanical energy holds true $$T+U=\dfrac{1}{2}mv^2+U=\text{constant}$$ differentiating with respect to position $x$ we get $$\dfrac{dT}{dx}=-\dfrac{dU}{dx}=F=ma$$ therefore by the same token one expects that kinetic energy can be defined up to an arbitrary additive constant such that $$T_c=\dfrac{1}{2}mv^2+c$$ where usually we prefer to set $c=0$ for simplicity. Is this fact right about kinetic energy?

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  • $\begingroup$ $T=\frac{1}{2}m\left(\frac{dx}{dt}\right)^2$ does not depend on $x$, so in your computation $\frac{dT}{dx}$ should be 0! See @Jon for the right way to do it. $\endgroup$
    – user154997
    Jun 13 '17 at 10:33
  • $\begingroup$ @LucJ.Bourhis Nope it does not. $$\dfrac{dT}{dx}=\dfrac{d[\dfrac{1}{2}mv^2]}{dx}=mv \dfrac{dv}{dx}=mv \dfrac{dv}{dt} \dfrac{dt}{dx}=mva\dfrac{1}{v}=ma=F$$ just as stated compactly in my question. $\endgroup$
    – Omar Nagib
    Jun 13 '17 at 10:43
  • $\begingroup$ That computation works only in one dimension… And really, in mechanics position and velocity have to be considered as independent variable. I mean you can independently fix the initial position and the initial velocity e.g. $\endgroup$
    – user154997
    Jun 13 '17 at 10:49
  • $\begingroup$ Mathematics aside, isn't it the case that in the rest frame of a particle, the kinetic (resulting from motion) energy of the particle is zero? This isn't arbitrary is it? $\endgroup$
    – Hal Hollis
    Jun 13 '17 at 12:25
  • $\begingroup$ @OmarNagib One must be careful when applying the chain rule: you cannot just insert differentials left and right. To start with, one must define what variable is $v$ function of (say, $t$); when so, one must then make sure that such function is invertible for the position in any point of the domain (it usually isn't) and so is its inverse derivatives. $\endgroup$
    – gented
    Jun 13 '17 at 13:13
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That's right but be careful about that constant that takes an interesting value from special relativity. Anyhow, I prefer this other approach. Consider $$ m\frac{d{\bf v}}{dt}={\bf F} $$ and multiply both members of this equation by ${\bf v}$. You will get $$ m{\bf v}\cdot\frac{d{\bf v}}{dt}={\bf F}\cdot{\bf v} $$ that means $$ \frac{d}{dt}\left(\frac{1}{2}mv^2\right)={\bf F}\cdot{\bf v}. $$ You can integrate in time obtaining $$ \frac{1}{2}mv^2+c=\int{\bf F}\cdot{\bf v}dt. $$ The constant is generally fixed by the problem at hand, e.g. by computing the work of the force on the right hand side.

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    $\begingroup$ The work is by definition the integral of the right hand side along a path, it isn't the "primitive" of the differential form. As such, one obtains eventually that $W_{\gamma} = T(B) - T(A)$ and no such constant ever appears in the definition of kinetic energy, the latter being defined exactly as the quadratic form $T(x,y,z) = 1/2 m (\dot{x}^2+\dot{y}^2+\dot{z}^2)$. $\endgroup$
    – gented
    Jun 13 '17 at 10:44
  • $\begingroup$ I have just integrated a differential equation aside from definitions. $\endgroup$
    – Jon
    Jun 13 '17 at 11:30
  • $\begingroup$ Yes, my objection is that the right hand side does not equal the work: it does so only if you integrated along a path. $\endgroup$
    – gented
    Jun 13 '17 at 12:54
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My answer would be yes; in the same frame of reference the work done on the mass $m$ will be the same for every observer:

$$W_{1-2}=\int_{t_1}^{t_2}\vec{F}.\vec{v}dt= \int_{t_1}^{t_2}\frac{d}{dt}\left(\frac{mv^2}{2}\right)dt=\frac{mv_2^2}{2} - \frac{mv_1^2}{2} = \Delta E_{kin}$$

so therfore we can a time-independent constant to the function in the second integral - $\frac{mv^2}{2}$ - without changing the result. And because kinetic energy can be definied as this function, the kinetic energy is definied up to a constant.

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